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Maslowich
2 years ago
7

During the day, susan notices that the wind is blowing onshore at the beach. What is this called? land breeze land breeze sea br

eeze sea breeze monsoon monsoon global wind
Physics
1 answer:
In-s [12.5K]2 years ago
8 0

The correct option is (b) Sea breeze

During the day, susan notices that the wind is blowing onshore at the beach called sea breeze.

What is sea breeze?

  • Any wind that flows from a big body of water onto or onto a landmass is called a sea breeze or an onshore breeze.
  • Sea breezes form as a result of changes in air pressure brought on by the different heat capacities of water and dry land. Sea breezes are therefore more confined than prevailing winds.
  • A sea wind is frequently seen along coasts after sunrise because land warms up far more quickly than water does when exposed to solar radiation.
  • The sea wind front is significant because it can serve as a catalyst for afternoon thunderstorms and provide welcome cooling along the coast.

Learn more about the sea breeze with the help of the given link:

brainly.com/question/13015619

#SPJ4

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Brainliest for correct answer! :)
Katena32 [7]

Answer:

1. 300.15

2. 873.15

3. 221.95

1. -243.15

2. -210.85

3. 295.17

5 0
3 years ago
An astronaut is on the moon. He drops a hammer from a height of 3.2metres and it takes 2.0 seconds to reach the lunar landscape.
Anvisha [2.4K]

Answer:

1/6 m/s^2      ( about 1/6th gravity of Earth ( 9.81 m/s^2)

Explanation:

Displacement =  yo  +  vo t  - 1/2 a t^2

      -  3.2          = 0     +  0     - 1/2 a(2.0)^2

      -     3.2       =                -2a

             a = 3.2 / 2 = 1.6 m/s^2

6 0
2 years ago
What is an organism's specific role in an ecosystem
Harman [31]
<span>Niche - </span><span>An organism's particular role in an ecosystem, or how it makes its living.</span>
7 0
3 years ago
Read 2 more answers
An object experiences a net acceleration to the left. Which of the following statements about this object are true? There may be
dedylja [7]

Answer:

  1. When an object experiences acceleration to the left, the net force acting on this object will also be to the left.
  2. If the mass of the object was doubled, it would experience an acceleration of half the magnitude

Explanation:

When an object experiences acceleration to the left, the net force acting on this object will also be to the left.

From Newton's second law of motion, the acceleration of the object is given as;

a = ∑F / m

a = -F / m

The negative value of "a" indicates acceleration to the left

where;

∑F is the net force on the object

m is the mass of the object

At a constant force, F = ma ⇒ m₁a₁ = m₂a₂

If the mass of the object was doubled, m₂ = 2m₁

a₂ = (m₁a₁) / (m₂)

a₂ = (m₁a₁) / (2m₁)

a₂ = ¹/₂(a₁)

Therefore, the following can be deduced from the acceleration of this object;

  1. When an object experiences acceleration to the left, the net force acting on this object will also be to the left.
  2. If the mass of the object was doubled, it would experience an acceleration of half the magnitude

6 0
3 years ago
Using a scale diagram, calculate the resultant force acting on a sailing boat when an easterly wind provides 2, point, 50, k, N,
EastWind [94]

Answer:

F = 3.6 kN, direction is 9.6º to the North - East

Explanation:

The force is a vector, so one method to find the solution is to work with the components of the vector as scalars and then construct the resulting vector.

Let's use trigonometry to find the component of the forces, let's use a reference frame where the x-axis coincides with the East and the y-axis coincides with the North.

Wind

X axis

          F₁ = 2.50 kN

Tide

         cos 30 = F₂ₓ / F₂

         sin 30 = F_{2y} / F₂

          F₂ₓ = F₂ cos 30

         F_{2y} = F₂ sin 30

         F₂ₓ = 1.20cos 30 = 1.039 kN

         F_{2y} = 1.20 sin 30 = 0.600 kN

the resultant force is

X axis

        Fₓ = F₁ₓ + F₂ₓ

        Fₓ = 2.50 +1.039

        Fₓ = 3,539 kN

        F_y = F_{2y}

        F_y = 0.600

to find the vector we use the Pythagorean theorem

         F = \sqrt{F_x^2 +F_y^2}

         F = \sqrt{ 3.539^2 + 0.600^2 }

         F = 3,589 kN

the address is

         tan θ = F_y / Fₓ

         θ = tan⁻¹ \frac{F_y}{F_x}

         θ = tan⁻¹  \frac{0.6}{3.539}0.6 / 3.539

         θ = 9.6º

the resultant force to two significant figures is

         F = 3.6 kN

the direction is 9.6º to the North - East

7 0
3 years ago
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